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    (Original post by SM-)
    Also for the multiple choice with integrals and also the one with A and B being max or something can someone show how they got they're answers?

    Also for the last multiple choice what was option A and option B and was it asking which one is false?
    here's wot i put earlier about the integral question...
    my thoughts for question 1H:
    1) y = a - x^2
    2) y = x^4 - a

    set y=0 and find x-intercepts:
    1) root(a), -root(a)
    2) 4throot(a), -4throot(a)

    integrate both between limits (x-intercepts):
    1) [ax - x^3/3] = (a^1.5 - (a^1.5)/3) - (-a^1.5 - (-a^1.5)/3) = 2a^1.5 - 2a^1.5/3 = 4/3 * a^1.5 = 4a.root(a)/3
    2) [x^5/5 - ax] = ((a^1.25)/5 - a^1.25) - ((-a^1.25)/5 - (-a^1.25)) = 2a^1.25/5 - 2a^1.25 = -8/5 * a^1.25 = -8a.fourthroot(a)/5

    set the two areas equal to each other:
    4/3 * a^1.5 = 8/5 * a^1.25
    a^0.25 = 6/5 a = (6/5)^4
    only one answer to do with (6/5)^4
    not sure about question 1I about max(ax+by)...
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    (Original post by 0x3bfc9a1)
    I am not so sure what you mean by (214-96)*6>214. For Q4 you didn't get alpha's simplest form, I don't know how you are supposed to get the area correct since you will need the angle to calculate the sector area.
    Well i just used alpha=arctan sqrt(3)/3 instead of pi/6. In question 2, A^nB^m(x)=2^n3^mx+2^n3^m-1 so there are 214-96=118 sums (k=118) and each x coefficient is greater or equal than 2^13^1=6 because m_i and n_i were positive. But 6*118>214 so is not possible.
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    (Original post by riemmanmath)
    Well i just used alpha=arctan sqrt(3)/3 instead of pi/6. In question 2, A^nB^m(x)=2^n3^mx+2^n3^m-1 so there are 214-96=118 sums (k=118) and each x coefficient is greater or equal than 2^13^1=6 because m_i and n_i were positive. But 6*118>214 so is not possible.
    214x - 96 was the sum of many (A^nB^m(x)), not just one.
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    (Original post by boombox111)
    214x - 96 was the sum of many (A^nB^m(x)), not just one.
    Precisely, what is your point? You have that all the A^nB^m(x) are lineal polynomial of the form ax+a-1 so there have to be 214-96 sums of A^nB^m(x) to get the polynomial 214x+96.
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    (Original post by riemmanmath)
    Precisely, what is your point? You have that all the A^nB^m(x) are lineal polynomial of the form ax+a-1 so there have to be 214-96 sums of A^nB^m(x) to get the polynomial 214x+96.
    so it's impossible.
    yeah?!
    i think we're arguing the same point.
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    (Original post by boombox111)
    so it's impossible.
    yeah?!
    i think we're arguing the same point.
    Yes it is impossible, but I just was saying that by a wide margin (i.e 214x+120 wouldnt be possible either)
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    How many marks would be the last two parts of each of 2 3 and 5 as well as the part about explaining integral in question 3?
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    How did people do Q4? I got the angle as pi/3??
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    Here I will like to explain some questions I found many ppl got so confused:
    1A come on you are supposed to find the product not the sum
    The MC with ax+by: Plug in a and b both equal to 1, then only option C will remain
    The so complicated graph of (x-1)^2-cos(pix): Plug in (0,0) then D and E eliminated
    Differentiate to find out that the graph has one local max only hence B and C eliminated
    The one with X^2+1: Since there are only even powers we can let y=x^2 then it should be crystal clear
    x(n) and pi(n): Note if pi(n)=1, there is only one prime factor hence this must be prime number hence A is wrong
    Q2: (only about the 214x+92) that one was trivial using brute force since coefficient of x must be multiple of 6. Note that m and n cannot be 0
    because the question says m and n are positive integers
    Q3v) Note whole graph is symmetrical along x=alpha by the definition of bilateral functions.

    Feel free to ask xp
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    (Original post by Mystery.)
    How did people do Q4? I got the angle as pi/3??
    I don't really remember the question tbh, any idea when the paper will be put up?
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    (Original post by LaserRanger)
    Here I will like to explain some questions I found many ppl got so confused:
    1A come on you are supposed to find the product not the sum
    The MC with ax+by: Plug in a and b both equal to 1, then only option C will remain
    The so complicated graph of (x-1)^2-cos(pix): Plug in (0,0) then D and E eliminated
    Differentiate to find out that the graph has one local max only hence B and C eliminated
    The one with X^2+1: Since there are only even powers we can let y=x^2 then it should be crystal clear
    x(n) and pi(n): Note if pi(n)=1, there is only one prime factor hence this must be prime number hence A is wrong
    Q2: (only about the 214x+92) that one was trivial using brute force since coefficient of x must be multiple of 6. Note that m and n cannot be 0
    because the question says m and n are positive integers
    Q3v) Note whole graph is symmetrical along x=alpha by the definition of bilateral functions.

    Feel free to ask xp
    The question 2 I got very mad afterwards (although I solved it) because I inmediatly supposed 214 would be divisible by 6 but isn't...xD. I also feel that the problem should say m,n nonnegative to make it more interesting.
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    (Original post by riemmanmath)
    The question 2 I got very mad afterwards (although I solved it) because I inmediatly supposed 214 would be divisible by 6 but isn't...xD. I also feel that the problem should say m,n nonnegative to make it more interesting.
    Why not define its inverse function hence allowing negative m,n right xd
    that will be fun enough
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    (Original post by goofyygoober)
    Hi, you have mentioned that the exam paper will be online on Friday. Is that on oxford maths department website?
    The 2016 paper is now up on the Oxford Maths website.
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    (Original post by OxfordMathsDept)
    The 2016 paper is now up on the Oxford Maths website.
    Hi when will the solutions be published?
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    (Original post by OxfordMathsDept)
    The 2016 paper is now up on the Oxford Maths website.
    Thank you! When will the solutions be available?
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    (Original post by Jacob2215)
    Thank you! When will the solutions be available?
    Solutions will go up online in January.
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    So, who's volunteering to make a provisional mark scheme?
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    The answers should be dbedabdecb
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    (Original post by 0x3bfc9a1)
    The answers should be dbedabdecb
    I disagree with your answer to question C of the MC.

    The answer to question C was 'a', which is that c is greater than zero.
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    (Original post by 0x3bfc9a1)
    The answers should be dbedabdecb
    shouldnt 1c be a
 
 
 
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